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Math Help - intergration (double)

  1. #1
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    intergration (double)

    evaluate the intergral

    3 2
    (integral sign) (intergral sign) (2y3+6xy+x2+1)dydx
    0 1


    so its like 0<x<3 and 1<y<2 (i think)


    any help?!
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  2. #2
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    Quote Originally Posted by turion645344 View Post
    evaluate the intergral

    3 2
    (integral sign) (intergral sign) (2y3+6xy+x2+1)dydx
    0 1


    so its like 0<x<3 and 1<y<2 (i think)


    any help?!
    \int_0^3 \int_1^2 2y^3 + 6xy + x^2 +1 \ dy \ dx

    \int_0^3 \frac{1}{2}y^4+3xy^2+x^2y+y \big|_1^2 \ dx

    \int_0^3 8+12x+2x^2+2 - \frac{1}{2} - 3x - x^2 - 1 \ dx

    You can do it fron there.
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  3. #3
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    when you take them away, i get

    3
    (intergration sign) x2+9x+8.5 dx

    0

    when
    3
    (1/3x3 + 9/2x2 + 8.5x)
    0

    and then 9+40.5+25.5 so an answer of 75.
    Last edited by turion645344; July 11th 2007 at 07:47 AM.
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  4. #4
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    the answer of 75. is this correct???
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  5. #5
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    Quote Originally Posted by turion645344 View Post
    the answer of 75. is this correct???
    Yes
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